Sunday, March 8, 2020

H C Verma solutions, Laws of Thermodynamics, EXERCISES, Q-11 to Q-22, Chapter-26, Concepts of Physics, Part-II

Laws of Thermodynamics

EXERCISES, Q-11 to Q-22

   11. A gas is taken through a cyclic process ABCA as shown in figure (26-E6). If 2.4 cal of heat is given in the process, what is the value of J?
The figure for Q-11
   

Answer:  From the laws of thermodynamics,

ΔQ = ΔU +ΔW, (Here ΔU =0, for a cyclic process).

→ΔQ = ΔW =ΔWₐᵦ+ΔWᵦ₍+ΔW₍ₐ 

{Area under the curve and volume graph}  

→(2.4 cal)*J = 0 + {½(200x10⁻⁶*100x10³)+(200x10⁻⁶*100x10³)} - (200x10⁻⁶*100x10³)

→(2.4 cal)*J =(3/2 -1)*(20) =10 Joules

→J = 10/2.4 = 4.17 J/cal 

 




  12. A substance is taken through a process abc as shown in figure (26-E7). If the internal energy of the substance increases by 5000 J and heat of 2625 cal is given to the system, calculate the value of J.
(26-E7)The figure for Q-12
   


Answer:  ΔQ =ΔU +ΔW 

Here ΔQ = 2625 cal =2625*J joules

ΔU = 5000 J

ΔW =ΔWₐᵦ +ΔWᵦ₍

     =(0.05-0.02)*200x10³ + 0 joules

     =6000 J

2625*J =5000 +6000 =11000

→J =11000/2625 =4.19 J/cal.

   



  13. A gas is taken along the path AB as shown in figure (26-E8). If 70 cal of heat is extracted from the gas in the process, calculate the change in the internal energy of the system.
The figure for Q-13
   


Answer:  Work done by the system,

ΔW =-(250-100)x10⁻⁶*{200 +½(500-200)x10³ J

→ΔW =-150*350*10⁻³ J = -52.5 J

ΔQ =-70 cal =-70*4.186 J

So, ΔU =ΔQ -ΔW =-70*4.186-(-52.5) J

           =241 J =Change in internal energy.   




 

  14. The internal energy of a gas is given by U = 1.5pV. It expands from 100 cm³ to 200 cm³ against a constant pressure of 1.0x10⁵ Pa. Calculate the heat absorbed by the gas in the process.   


Answer:  ΔQ =ΔU+ΔW

ΔW =pV =1.0x10⁵*(200-100)x10⁻⁶ J

=10 J

Change in internal energy

ΔU =1.5*1.0x10⁵*(200-100)x10⁻⁶ J 

=15 J

Heat absorbed ΔQ =10 J +15 J =25 J.



 

  15. A gas is enclosed in a cylindrical vessel fitted with a frictionless piston. The gas is slowly heated for some time. During the process, 10 J of heat is supplied and the piston is found to move out 10 cm. Find the increase in the internal energy of the gas. The area of the cross-section of the cylinder = 4 cm² and the atmospheric pressure = 100 kPa.    


Answer: Work done by the gas =p*dV

→ΔW=p*dV=100x10³*(4x10⁻⁴*10/100)

=4 J.
Heat given, ΔQ = 10 J.
Change in internal energy =ΔU =?
From the first law of thermodynamics,  

ΔQ =ΔU +ΔW

→10 = ΔU + 4

→ΔU = 10 - 4 = 6 J.   





 

  16. A gas is initially at a pressure of 100 kPa and its volume is 2.0 m³. Its pressure is kept constant and the volume is changed from 2.0 m³ to 2.5 m³. Its volume is now kept constant and the pressure is increased from 100 kPa to 200 kPa. The gas is brought back to its initial state, the pressure varying linearly with its volume. (a) Whether the heat is supplied to or extracted from the gas in the complete cycle? (b) How much heat was supplied or extracted?   


Answer:  At initial state A,

p = 100 kPa, V = 2.0 m³

At 2nd state B, p=100 kPa, V= 2.5 m³.

Hence dV =0.5 m³
Diagram for Q-16

Hence work done (for process AB)

= p*dV 

=(100 kPa)*(0.5 m³)

=100x1000*0.5 J

=50000 J 

At the state C, p=200 kPa, V=2.5 m³

So for process BC, dV =0, Thus here work done = zero. 

For the process CA,

dV = -0.5 m³. Work done = area under CA and V axis.

=-(100x1000*0.5+½*100x1000*0.5) J

=-1.5*50000 J

=-75000 J

Net work-done in the process ABCA,

ΔW =50000+0-75000 =-25000 J

For a cyclic process ΔU =0.

Hence from,

ΔQ =ΔU+ΔW

→ΔQ = 0 -25000 J =-25000 J.


(a) A negative sign shows that heat is extracted from the gas in the complete cycle.


(b) The amount of heat extracted =25000 J.     




 

  17. Consider the cyclic process ABCA, shown in figure (26-E9), performed on a sample of 2.0 mole of an ideal gas. A total of 1200 J of heat is withdrawn from the sample in the process. Find the work done by the gas during the part BC.
The figure for Q-17
   

Answer:  For the complete cycle, ΔQ = -1200 J, ΔU = 0. From the first law of thermodynamics,

ΔQ =ΔU+ΔW

→-1200 = 0+ ΔW

→ΔW = -1200 J = work done in the whole cycle.

During the process of CA, the volume does not change. So no work is done in this process.

During the process, AB the temperature is proportional to the volume hence here the pressure must be constant. 

Now pV = nRT
At A, pVₐ = nR(300)
At B, pVᵦ = nR(500)
Hence, pVᵦ-pVₐ =nR(500-300)
→p(Vᵦ-Vₐ) =nR*200
→p*𐊅V =2*8.3*200
→Work done in the process AB=3320 J. 

Now the total work done in the process AB, BC, and CA = 3320 +Wᵦ₍+0
=Wᵦ₍+3320 J.
But the work done in the whole cycle =-1200 J. Hence,
Wᵦ₍ +3320 =-1200
→Wᵦ₍ =-3320-1200 =4520 J. 

 


 

  18. Figure (26-E10) shows the variation in the internal energy U with the volume V of 2.0 mole of an ideal gas in the cyclic process abcda. The temperature of the gas at b and c are 500 K and 300 K respectively. Calculate the heat absorbed by the gas during the process.
The figure for Q-18
   


Answer:  During the cyclic process, 𐊅U = 0. Hence 𐊅Q = 𐊅W. In the given picture, the volume during the processes bc and da is constant. Hence no work is done during these processes. During the processes ab and cd, the temperature is constant and these are isothermal processes.

The temperature at a and b, T = 500 K.

The temperature at c and d, T'=300 K.

Work done during an isothermal process is given as, W = nRT*ln(V₂/V₁).

The work done during the process ab is W= nRT*ln(2V₀/V₀) =nRT*ln(2), 

Similarly, the work done during the process cd, W' = nRT'*ln(V₀/2V₀) =nRT'*ln(1/2).

So the total work done during the cyclic process abcda, 𐊅W = W+W'

→𐊅W =nRT*ln(2) +nRT'*ln(1/2)

=nRT*ln(2) +nRT'*ln(1) -nRT'*ln(2)

=nR*ln(2){T-T'} +nRT'*0
{Since ln(1) =0}
=nR*ln(2){500-300}
=2*8.3*0.693*200
=2300 J.

As we have seen in the beginning, here 𐊅Q =𐊅W

The total heat absorbed during this cyclic process =𐊅Q =𐊅W =2300 J.

  


 


  19. Find the change in the internal energy of 2 kg of water as it is heated from 0°C to 4°C. The specific heat capacity of water is 4200 J/kg-K and its densities at 0°C and 4°C are 999.9 kg/m³ and 1000 kg/m³ respectively. Atmospheric pressure =10⁵ Pa.   


Answer:  𐊅Q = ms*𐊅T

=2*4200*4 =33600 J.

Pressure, p = 10⁵ Pa.

Change in volume, 

𐊅V =2/1000-2/999.9 =2.0x10⁻⁷ m³

Hence 𐊅W =p*𐊅V =10⁵*2.0x10⁻⁷ J

               =0.02 J.

From the first law of thermodynamics, 𐊅Q =𐊅W+𐊅U

→𐊅U =𐊅Q -𐊅W =(33600 -0.02) J.

= The change in the internal energy. 


 


  20. Calculate the increase in the internal energy of 10 g of water when it is heated from 0°C to 100°C and converted into steam at 100 kPa. The density of steam = 0.6 kg/m³. Specific heat capacity of water = 4200 J/kg-°C and the latent heat of vaporization of water = 2.5x10⁶ J/kg.   


Answer:  Heat given to water in the process 𐊅Q =Heat given to water to raise the temperature from 0°C to 100°C + Heat given to convert the water into steam

=ms*𐊅T +mL

=(10/1000)*4200*100 +(10/1000)*2.5x10⁶

=4200 +25000 

=29200 J

The density of water varies with temperature but the density of steam is much less. Hence the change in the volume of water from 0°C to 100°C is negligible in comparison to the steam. Taking the density of water =1000 kg/m³, volume =(10/1000)/1000 m³ 

=1x10⁻⁵ m³ 

The density of steam at 100°C (given) =0.6 kg/m³, volume =(10/1000)/0.6 m³

=0.01666 m³

Chang in volume, 𐊅V =(0.01666 -1x10⁻⁵) m³

Pressure, p = 100 kPa =1x10⁵ Pa

The work done, 𐊅W =p*𐊅V

=1x10⁵*(0.01666-1x10⁻⁵) J

=1665 J.

If 𐊅U = change in internal energy, then from the first law of thermodynamics,

𐊅Q =𐊅W +𐊅U

→𐊅U =𐊅Q -𐊅W =29200-1665 = 27535 J =2.75x10⁴ J.       



 

 

 

  21. Figure (26E-11) shows a cylindrical tube of volume V with adiabatic walls containing an ideal gas. The internal energy of this ideal gas is given by 1.5nRT. The tube is divided into two equal parts by a fixed diathermic wall. Initially, the pressure and the temperature are p₁, T₁ on the left and p₂, T₂ on the right. The system is left for sufficient time so that the temperature becomes equal on the two sides. (a) How much work has been done by the gas on the left part? (b) Find the final pressure on the two sides. (c) Find the final equilibrium temperature. (d) How much heat has flown from the gas on the right to the gas on the left?
The figure for Q-21
    


Answer:  (a) Since the diathermic wall is fixed, there is no change in the volumes of the gases on either side. So zero work is done by the gas on the left part.   


Let us first answer the part (c)
(c) Let the final temperature on both parts = T. n₁ =Number of moles on the left part. n₂ =Number of moles on the right part. 
  
Now, p₁(V/2) =n₁RT₁, p₂(V/2)=n₂RT₂,
→n₁ =p₁V/2RT₁ and n₂ =p₂V/2RT₂
Total number of moles, n =n₁+n₂
=(p₁V/2RT₁ +p₂V/2RT₂)
Internal energy is given as 1.5nRT, hence the internal energy on the left part, U₁=1.5n₁RT₁,
On the right part, U₂ =1.5n₂RT₂
The total internal energy of the gas initially, U =U₁+U₂ 
=1.5n₁RT₁ +1.5n₂RT₂ 
=1.5(p₁V/2RT₁)RT₁+1.5(p₂V/2RT₂)RT₂  =0.75p₁V+0.75p₂V
=0.75V(p₁+p₂) -------- (i)
Internal energy when equilibrium is reached =1.5nRT
=1.5(p₁V/2RT₁ +p₂V/2RT₂)RT --(ii) 

 Since no work is done by the gases on either part and no heat is transferred due to adiabatic walls, the internal energy will remain the same. Equating (i) and (ii),

1.5(p₁V/2RT₁ +p₂V/2RT₂)RT=0.75V(p₁+p₂) 
→0.75(p₁/2T₁+p₂/2T₂)T=0.75(p₁+p₂)
→T =(p₁+p₂)/(p₁/T₁+p₂/T₂)
=T₁T₂(p₁+p₂)/(p₁T₂+p₂T₁)
=T₁T₂(p₁+p₂)/𝜆 =Final equilibrium temperature.
where 𝜆 =p₁T₂+p₂T₁

 Now part (b)  
Assume that final pressure on left part =p, on right part =p'  
For the left part,
p₁(V/2)/T₁ =n₁R =p(V/2)/T 
→p =p₁T/T₁
→p =p₁T₁T₂(p₁+p₂)/𝜆T₁ 
{Putting the value of T)
→p =p₁T₂(p₁+p₂)/𝜆 

 Similarly, for the right side the final pressure p' =p₂T/T₂ 
→p' =p₂T₁T₂(p₁+p₂)/𝜆T₂
→p' =p₂T₁(p₁+p₂)/𝜆  

 (d) Since no work is done by the gas on the right side, the change in the internal energy of the gas on the right side will be the heat flown from the right side.
The change in the internal energy of the gas on the right side =1.5n₂R(T₂-T) 

 =1.5(p₂V/2RT₂)R{T₂-T₁T₂(p₁+p₂)/𝜆}    =(3/4)(p₂V){1-T₁(p₁+p₂)/(p₁T₂+p₂T₁)}
Putting the value of 𝜆 above 
=(3/4)(p₂V){(p₁T₂+p₂T₁-p₁T₁-p₂T₁)/𝜆} 
=(3/4)(p₂V){(p₁T₂-p₁T₁)/𝜆} 
=3p₁p₂(T₂-T₁)V/4𝜆    


  22. An adiabatic vessel of total volume V is divided into two equal parts by a conducting separator. The separator is fixed in this position. The part on the left contains one mole of an ideal gas (U = 1.5nRT) and the part on the right contains two moles of the same gas. Initially, the pressure on each side is p. The system is left for sufficient time so that a steady-state is reached. Find (a) the work done by the gas on the left part during the process. (b) The temperature on the two sides in the beginning, (c) the final common temperature reached by the gases. (d) the heat given to the gas in the right part and (e) the increase in the internal energy of the gas in the left part.    


Answer:  (a) Since the separator is fixed in position the volume of the gas does not change on the left part, hence the work done by the gas in the left part during the process is zero.  


(b) Let the temperatures in the beginning and at the left and right parts be T₁, T₂, respectively. 

So for the left side, p(V/2)=(1 mol)RT₁

T₁ =pV/(2 mol)R

For the right side, p(V/2)=(2 mol)RT₂

T₂ =pV/(4 mol)R


(c) Let the final temperature =T 

Initial internal energy, U =U₁+U₂

=1.5n₁RT₁+1.5n₂RT₂ 

=1.5*1*R*pV/2R +1.5*2*R*pV/4R

=3pV/4 +3pV/4

=6pV/4

=1.5pV  

The final internal energy =1.5(3 mol)RT

=(4.5 mol)RT    

Since no work is done by the gas as no volume change, the change in internal energy will be the heat given. Since walls of the vessel are adiabatic no heat is exchanged. Thus the internal energy does not change.

Equating the initial and final internal energies we get,
(4.5 mol)RT =1.5pV
►T =1.5pV/(4.5 mol)R
      = pV/(3 mol)R


(d) The heat given to the gas in the right part = Increase in the internal energy of the right part,

=1.5n₂R(T-T₂)
=1.5*2*R(pV/3R -pV/4R)
=3(pV/3 -pV/4)
=3(pV/12)
= pV/4

(e) Since the walls are adiabatic, the heat given to the right part comes from the left part. Hence for the left part ΔQ = -pV/4
Since for the left part ΔW=0, ΔQ =ΔU
→ΔU= increase in the internal energy of the left part =ΔQ = -pV/4. 


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Links to the Chapters



CHAPTER- 20 - Dispersion and Spectra


CHAPTER- 19 - Optical Instruments

CHAPTER- 18 - Geometrical Optics



CHAPTER- 17 - Light Waves




CHAPTER- 14 - Fluid Mechanics



CHAPTER- 13 - Fluid Mechanics


CHAPTER- 12 - Simple Harmonic Motion








CHAPTER- 11 - Gravitation




CHAPTER- 10 - Rotational Mechanics




CHAPTER- 9 - Center of Mass, Linear Momentum, Collision


CHAPTER- 8 - Work and Energy

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CHAPTER- 7 - Circular Motion

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CHAPTER- 4 - The Forces

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CHAPTER- 3 - Kinematics - Rest and Motion

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CHAPTER- 2 - "Physics and Mathematics"

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